Question

A bucket is $40cm$ in diameter at the top and $28cm$ in diameter at the bottom. Find the capacity of the bucket in litres, if it is $21cm$ deep. Also, find the cost of tin sheet used in making the bucket, if the cost of tin is Rs $1.50$ per sq dm.

Answer 1

Hint:
Here, we will find the capacity of the bucket or we can say volume of the bucket by using the volume of the truncated cone formula as the shape of the bucket is like a truncated cone. To find the cost of a tin sheet we will find the surface area of the bucket and for that we will find the slant height of the bucket.

Formula Used:
We will use the following formulas:
1) Volume of truncated cone $ = \dfrac{{\pi h}}{3} \times \left( {{r_1}^2 + {r_1}{r_2} + {r_2}^2} \right)$ where, $h$ is the height of the bucket, ${r_1}$ is the radius of top circular shape of the bucket, ${r_2}$ is the radius of bottom circular shape of the bucket.
2) Slant height of the bucket $l = \sqrt {{{\left( {{r_1} - {r_2}} \right)}^2} + {h^2}} $
3) Total Surface Area of the bucket $ = \pi l\left( {{r_1} + {r_2}} \right) + \pi {r_2}^2$

Complete step by step solution:
As given diameter of bucket at top and bottom is $40cm$ and $28cm$ respectively
So, radius of bucket at top and bottom will be half of its diameter as
${\text{radius}} = \dfrac{{{\text{diameter}}}}{2}$
So, we get our radius as
 ${r_1} = \dfrac{{40}}{2}$
$ \Rightarrow {r_1} = 20cm$
${r_2} = \dfrac{{28}}{2}$
$ \Rightarrow {r_2} = 14cm$
Next, height of the bucket is given as $h = 21cm$.
Substituting all the values in the formula Volume of truncated cone $ = \dfrac{{\pi h}}{3} \times \left( {{r_1}^2 + {r_1}{r_2} + {r_2}^2} \right)$, we get,
Volume of truncated cone $ = \dfrac{{\pi \times 21}}{3}\left( {{{\left( {20} \right)}^2} + 20 \times 14 + {{\left( {14} \right)}^2}} \right)$
Applying the exponent on the terms, we get
$ \Rightarrow $ Volume of truncated cone $ = 7\pi \left( {400 + 280 + 196} \right)$
Adding the terms, we get
$ \Rightarrow $ Volume of truncated cone $ = 7 \times \dfrac{{22}}{7} \times 876$
Multiplying the terms, we get
$ \Rightarrow $ Volume of truncated cone $=19272$ cm3
Next, we have to find the volume (capacity) in litres so we will divide our answer from 1000, we get
$ \Rightarrow $ Volume of truncated cone $ = \dfrac{{19272}}{{1000}}$ litres
So, the volume of the truncated cone is $19.272$ litres.
Next, we find the slant height of the bucket by using $l = \sqrt {{{\left( {{r_1} - {r_2}} \right)}^2} + {h^2}} $ formula. Therefore, we get
 $l = \sqrt {{{\left( {20 - 14} \right)}^2} + {{21}^2}} $
Subtracting and applying the exponent, we get
$ l = \sqrt {{6^2} + 441} \\
  l = \sqrt {36 + 441} \\ $
Adding the terms, we get
$ l = \sqrt {477} \\
  l = 21.84cm \\ $
Next, we will find the Total Surface area of the bucket by using $ = \pi l\left( {{r_1} + {r_2}} \right) + \pi {r_2}^2$ formula and get,
Total Surface area $ = \pi \times 21.84\left( {20 + 14} \right) + {14^2}$
Adding the terms and applying the exponent, we get
$ \Rightarrow $ Total Surface area $ = \dfrac{{22}}{7} \times 21.84 \times 34 + 196$
On simplifying the expression, we get
$ \Rightarrow $ Total Surface area $ = 2949.76c{m^2}$
Therefore required cost of the tin sheet at Rs $1.50$ per sq dm (per 100cm2) is
Required cost of the tin sheet $ = {\text{Rs}}\dfrac{{1.50 \times 2949.76}}{{100}}$
$ \Rightarrow $ Required cost of the tin sheet $ = {\text{Rs}}44.25$

Hence, capacity of Bucket is $19.272$ litres and cost of tin sheet used in making it is Rs $44.25$.

Note:
In this question we might make a mistake if we forget to change the unit in litres. Litre is the SI unit of Volume and it should be always expressed in the standard form only. Another mistake that we can make is when we find the cost of the tin sheet we directly multiply the Total Square Area with the cost of one unit of it. But this will give us the wrong answer because we are asked to find cost per sq dm and our area is in sq cm. So we need to change the unit of area to sq dm before finding the cost.